CBSE Class 12 Maths Question Paper 2026 Solved Code 65-1-3

SECTION A

Question 1 to 20 are multiple choice questions of 1 mark each.

1. If \( |\vec{a}| = 8, |\vec{b}| = 3 \) and \( |\vec{a} \times \vec{b}| = 12 \), then the value of \( |\vec{a} \cdot \vec{b}| \) [1 Mark]
(A) \( 6\sqrt{3} \)
(B) \( 8\sqrt{3} \)
(C) \( 12\sqrt{3} \)
(D) \( 3\sqrt{12} \)

Answer: (C) \( 12\sqrt{3} \)

Teacher's Note:
a) Use \( \sin\theta = \dfrac{|\vec{a}\times\vec{b}|}{|\vec{a}||\vec{b}|} = \dfrac{12}{24} = \dfrac{1}{2} \), so \( \cos\theta = \dfrac{\sqrt{3}}{2} \).
b) Then \( \vec{a}\cdot\vec{b} = |\vec{a}||\vec{b}|\cos\theta = 8 \times 3 \times \dfrac{\sqrt{3}}{2} = 12\sqrt{3} \).

2. The length of perpendicular drawn from the point (3, 4, 2) on the line \( \dfrac{x}{0} = \dfrac{y}{0} = \dfrac{z}{1} \) is [1 Mark]
(A) 2
(B) 9
(C) 5
(D) \( \sqrt{29} \)

Answer: (C) 5

Teacher's Note:
a) The given line is the z-axis, so the foot of perpendicular is (0, 0, 2).
b) Distance \( = \sqrt{3^2+4^2+0^2} = 5 \).

3. The feasible region of a linear programming problem with objective function Z = 5x + 7y is shown below : [1 Mark]

[Figure: A shaded feasible region bounded by points (0,2), (3,4), and (7,0), with the region enclosed between the axes and two line segments meeting at (3,4).]

The maximum value of Z - minimum value of Z is
(A) 8
(B) 29
(C) 35
(D) 43

Answer: (D) 43

Teacher's Note:
a) Evaluate Z at each corner point: (0,0) gives 0, (7,0) gives 35, (3,4) gives 43, (0,2) gives 14.
b) Maximum Z = 43, minimum Z = 0, so the difference is 43.

4. The degree of an objective function of a linear programming problem is [1 Mark]
(A) 0
(B) 1
(C) 2
(D) Any natural number

Answer: (B) 1

Teacher's Note:
a) The objective function of an LPP is always linear.
b) A linear function always has degree 1.

5. If \( \sin^{-1}x + \pi = y \), then [1 Mark]
(A) \( -\dfrac{\pi}{2} \le y \le \dfrac{\pi}{2} \)
(B) \( -\dfrac{3\pi}{2} \le y \le -\dfrac{\pi}{2} \)
(C) \( \dfrac{\pi}{2} \le y \le \dfrac{3\pi}{2} \)
(D) \( 0 \le y \le \pi \)

Answer: (C) \( \dfrac{\pi}{2} \le y \le \dfrac{3\pi}{2} \)

Teacher's Note:
a) The range of \( \sin^{-1}x \) is \( \left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right] \).
b) Adding \( \pi \) to all sides shifts the range to \( \left[\dfrac{\pi}{2}, \dfrac{3\pi}{2}\right] \).

6. If A = \( [a_{ij}]_{3 \times 3} \) is a scalar matrix then which of the following must be true ? [1 Mark]
(A) A must be a symmetric matrix.
(B) A must be a skew-symmetric matrix.
(C) A must be an identity matrix.
(D) A must be a null matrix.

Answer: (A) A must be a symmetric matrix.

Teacher's Note:
a) A scalar matrix has equal diagonal entries and zeros elsewhere, so it is always symmetric.
b) It need not be an identity matrix unless the scalar equals 1, and it need not be null.

7. Which of the following properties is/are true for two matrices of suitable orders ? [1 Mark]
(i) \( (A+B)' = A' + B' \)
(ii) \( (A-B)' = B' - A' \)
(iii) \( (AB)' = A'B' \)
(iv) \( (kAB)' = kB'A' \) (k is a scalar)
(A) (i) only
(B) (i), (ii) and (iii)
(C) (i) and (ii)
(D) (i) and (iv)

Answer: (D) (i) and (iv)

Teacher's Note:
a) \( (A+B)'=A'+B' \) is always true, but \( (A-B)'=A'-B' \), not \( B'-A' \).
b) \( (AB)'=B'A' \), not \( A'B' \); and \( (kAB)'=kB'A' \) is correct.

8. If \( \Delta_1 = \begin{vmatrix}1 & 0 & 0\\0 & 2 & 0\\0 & 0 & 3\end{vmatrix} \) and \( \Delta_2 = \begin{vmatrix}0 & 2 & 0\\1 & 0 & 0\\0 & 0 & 6\end{vmatrix} \), then [1 Mark]
(A) \( \Delta_1 = 2\Delta_2 \)
(B) \( \Delta_2 = -2\Delta_1 \)
(C) \( \Delta_1 = \Delta_2 \)
(D) \( \Delta_2 = -\Delta_1 \)

Answer: (B) \( \Delta_2 = -2\Delta_1 \)

Teacher's Note:
a) \( \Delta_1 = 1 \times 2 \times 3 = 6 \).
b) Expanding \( \Delta_2 \) gives \( -12 \), so \( \Delta_2 = -2\Delta_1 \).

9. One of the values of x for which \( \begin{vmatrix}\cos x & \sin x\\-\cos x & \sin x\end{vmatrix} = 1 \) is [1 Mark]
(A) 0
(B) \( \dfrac{\pi}{4} \)
(C) \( \dfrac{\pi}{3} \)
(D) \( \dfrac{\pi}{2} \)

Answer: (B) \( \dfrac{\pi}{4} \)

Teacher's Note:
a) Expanding gives \( \cos x \sin x + \sin x \cos x = \sin 2x \).
b) \( \sin 2x = 1 \) gives \( 2x = \dfrac{\pi}{2} \), so \( x = \dfrac{\pi}{4} \).

10. If A and B are symmetric matrics of same order, then (AB - BA) is a [1 Mark]
(A) Zero matrix
(B) Identity matrix
(C) Symmetric matrix
(D) Skew symmetric matrix

Answer: (D) Skew symmetric matrix

Teacher's Note:
a) \( (AB-BA)' = B'A' - A'B' = BA - AB = -(AB-BA) \).
b) Since it equals the negative of its transpose, it is skew symmetric.

11. The least value of f(x) = e^-x in [0, 3] is [1 Mark]
(A) \( e^{-3} \)
(B) -1
(C) 1
(D) \( -e^3 \)

Answer: (A) \( e^{-3} \)

Teacher's Note:
a) \( f(x)=e^{-x} \) is a decreasing function on [0,3].
b) So the least value occurs at the right end point, x = 3, giving \( e^{-3} \).

12. If \( \displaystyle\int \dfrac{3ax}{b^2+c^2x^2}\,dx = A \log|b^2+c^2x^2| + K \), then the value of A is [1 Mark]
(A) 3a
(B) \( \dfrac{3a}{2b^2} \)
(C) \( \dfrac{3a}{b^2c^2} \)
(D) \( \dfrac{3a}{2c^2} \)

Answer: (D) \( \dfrac{3a}{2c^2} \)

Teacher's Note:
a) The derivative of \( \log|b^2+c^2x^2| \) is \( \dfrac{2c^2x}{b^2+c^2x^2} \).
b) Comparing coefficients gives \( A = \dfrac{3a}{2c^2} \).

13. The value of \( \displaystyle\int_{-1}^{1} \dfrac{x^3}{x^2+2|x|+1}\,dx \) is [1 Mark]
(A) 0
(B) log 2
(C) 2 log 2
(D) \( \dfrac{1}{2}\log 2 \)

Answer: (A) 0

Teacher's Note:
a) The numerator \( x^3 \) is odd, and the denominator \( x^2+2|x|+1 \) is even.
b) So the integrand is an odd function, and the integral over a symmetric interval is 0.

14. The area bounded by the curve y = x|x|, x-axis and the ordinates x = -1 and x = 1 is given by [1 Mark]
(A) 0
(B) \( \dfrac{1}{3} \)
(C) \( \dfrac{2}{3} \)
(D) \( \dfrac{4}{3} \)

Answer: (C) \( \dfrac{2}{3} \)

Teacher's Note:
a) For \( x \ge 0 \), \( y=x^2 \); for \( x \lt 0 \), \( y=-x^2 \).
b) Area \( = \int_{-1}^0 x^2\,dx + \int_0^1 x^2\,dx = \dfrac{1}{3}+\dfrac{1}{3} = \dfrac{2}{3} \).

15. The integrating factor of differential equation \( R\dfrac{dx}{dy} + Px = Q \) where P, Q, R are functions of y is [1 Mark]
(A) \( e^{\int \frac{P}{Q}dy} \)
(B) \( e^{\int Pdy} \)
(C) \( e^{\int \frac{P}{R}dy} \)
(D) \( e^{\int \frac{P}{R}dx} \)

Answer: (C) \( e^{\int \frac{P}{R}dy} \)

Teacher's Note:
a) Rewrite as \( \dfrac{dx}{dy} + \dfrac{P}{R}x = \dfrac{Q}{R} \), the standard linear form in x.
b) The integrating factor is \( e^{\int (P/R)\,dy} \).

16. The order and degree of the differential equation : \( \dfrac{d}{dx}(y')^3 + (y')^3 = 1 \) respectively are where \( y' = \dfrac{dy}{dx} \) [1 Mark]
(A) 1, 3
(B) 2, 1
(C) 3, 1
(D) 3, 2

Answer: (B) 2, 1

Teacher's Note:
a) \( \dfrac{d}{dx}(y')^3 = 3(y')^2y'' \), so the equation becomes \( 3(y')^2y'' + (y')^3=1 \).
b) The highest derivative is \( y'' \) (order 2), appearing with power 1 (degree 1).

17. The value of p for which vectors \( \hat{i}+2\hat{j}+3\hat{k} \) and \( 2\hat{i}-p\hat{j}+\hat{k} \) are perpendicular to each other is [1 Mark]
(A) 0
(B) 1
(C) \( \dfrac{5}{2} \)
(D) \( -\dfrac{5}{2} \)

Answer: (C) \( \dfrac{5}{2} \)

Teacher's Note:
a) Perpendicular vectors have zero dot product: \( 2 - 2p + 3 = 0 \).
b) Solving gives \( p = \dfrac{5}{2} \).

18. The value of m for which the points with position vectors \( -\hat{i}-\hat{j}+2\hat{k} \), \( 2\hat{i}+m\hat{j}+5\hat{k} \) and \( 3\hat{i}+11\hat{j}+6\hat{k} \) are collinear, is [1 Mark]
(A) 8
(B) -8
(C) 2
(D) \( \dfrac{5}{2} \)

Answer: (A) 8

Teacher's Note:
a) Find direction vectors AB and AC and equate their ratios of components.
b) Solving \( \dfrac{m+1}{12}=\dfrac{3}{4} \) gives \( m=8 \).

Assertion - Reason Based Questions

Direction : Question numbers 19 and 20 are Assertion and Reason based questions carrying 1 mark each. Two statements are given, one labelled Assertion (A) and other labelled Reason (R).

Select the correct answer from the codes (A), (B), (C) and (D) as given below.

19. Assertion (A) : In an experiment of throwing an unbiased die, the probability of getting a prime number given that number appearing on the die being odd is \( \dfrac{2}{3} \). [1 Mark]
Reason (R) : For any two events A and B, \( P(A|B) = \dfrac{P(A \cup B)}{P(B)} \)
(A) Both Assertion (A) and Reason (R) are true and the Reason (R) is the correct explanation of the Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(C) Assertion (A) is true and Reason (R) is false.
(D) Assertion (A) is false and Reason (R) is true.

Answer: (C) Assertion (A) is true and Reason (R) is false.

Teacher's Note:
a) Odd numbers are 1, 3, 5; prime and odd numbers among them are 3, 5, giving \( P(A|B)=\dfrac{2/6}{3/6}=\dfrac{2}{3} \), so the Assertion is true.
b) The correct formula is \( P(A|B)=\dfrac{P(A \cap B)}{P(B)} \), not with union, so the Reason is false.

20. Assertion (A) : Lines given by x = py + q, z = ry + s and x = p'y + q', z = r'y + s' are perpendicular to each other when pp' + rr' = 1. [1 Mark]
Reason (R) : Two lines \( \vec{r}=\vec{a_1}+\lambda\vec{b_1} \) and \( \vec{r}=\vec{a_2}+\mu\vec{b_2} \) are perpendicular to each other if \( \vec{b_1}\cdot\vec{b_2}=0 \).
(A) Both Assertion (A) and Reason (R) are true and the Reason (R) is the correct explanation of the Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(C) Assertion (A) is true and Reason (R) is false.
(D) Assertion (A) is false and Reason (R) is true.

Answer: (D) Assertion (A) is false and Reason (R) is true.

Teacher's Note:
a) The direction ratios of the two lines are (p,1,r) and (p',1,r'); perpendicularity needs \( pp'+1+rr'=0 \), i.e. \( pp'+rr'=-1 \), not 1.
b) The Reason correctly states the general perpendicularity condition for two lines in vector form.

SECTION B

This section comprises Very Short Answer (VSA) type questions of 2 marks each.

21. (a) Simplify : \( \tan^{-1}\left(\dfrac{\cos 2x - \sin 2x}{\cos 2x + \sin 2x}\right) \), \( 0 \lt x \lt \dfrac{\pi}{4} \). [2 Marks]

Answer:
1. Dividing numerator and denominator by \( \cos 2x \) gives \( \tan^{-1}\left(\dfrac{1-\tan 2x}{1+\tan 2x}\right) = \tan^{-1}\left[\tan\left(\dfrac{\pi}{4}-2x\right)\right] \).
2. Since \( 0 \lt x \lt \dfrac{\pi}{4} \), this simplifies to \( \dfrac{\pi}{4} - 2x \).

Teacher's Note:
a) Recognise the tan subtraction formula pattern \( \dfrac{1-\tan A}{1+\tan A} = \tan\left(\dfrac{\pi}{4}-A\right) \).
b) Check the given domain to ensure the principal value branch is used correctly.

OR

(b) Evaluate : \( \tan\left(\sin^{-1}1 - \cos^{-1}\left(-\dfrac{1}{2}\right)\right) \) [2 Marks]

Answer:
1. \( \sin^{-1}1 = \dfrac{\pi}{2} \) and \( \cos^{-1}\left(-\dfrac{1}{2}\right) = \dfrac{2\pi}{3} \).
2. So the expression is \( \tan\left(\dfrac{\pi}{2}-\dfrac{2\pi}{3}\right) = \tan\left(-\dfrac{\pi}{6}\right) = -\dfrac{1}{\sqrt{3}} \).

Teacher's Note:
a) Recall standard values of inverse trigonometric functions at 1 and -1/2.
b) Simplify the angle first, then apply the tan value.

22. (a) Check whether function f(x) defined as
f(x) = \(\begin{cases}\dfrac{|x-3|}{2(x-3)}, & x \lt 3\\ \dfrac{x-6}{6}, & x \ge 3\end{cases}\) is continuous at x = 3 or not ? [2 Marks]

Answer:
1. For \( x \lt 3 \), \( |x-3| = -(x-3) \), so \( f(x) = \dfrac{-(x-3)}{2(x-3)} = -\dfrac{1}{2} \).
2. LHL \( = -\dfrac{1}{2} \), RHL \( = \displaystyle\lim_{x\to 3^+}\dfrac{x-6}{6} = -\dfrac{1}{2} \), and \( f(3) = \dfrac{3-6}{6} = -\dfrac{1}{2} \).
3. Since LHL = RHL = f(3), the function is continuous at x = 3.

Teacher's Note:
a) Always split the modulus function carefully according to the sign of x - 3.
b) Compare LHL, RHL and the function value at the point to conclude continuity.

OR

(b) If \( \sqrt{3}(x^2+y^2) = 4xy \), then find \( \dfrac{dy}{dx} \) at \( \left(\dfrac{1}{2}, \dfrac{\sqrt{3}}{2}\right) \). [2 Marks]

Answer:
1. Differentiating both sides: \( \sqrt{3}\left(2x+2y\dfrac{dy}{dx}\right) = 4\left(x\dfrac{dy}{dx}+y\right) \).
2. Solving gives \( \dfrac{dy}{dx} = \dfrac{2y-\sqrt{3}x}{\sqrt{3}y-2x} \).
3. At \( \left(\dfrac{1}{2}, \dfrac{\sqrt{3}}{2}\right) \), \( \dfrac{dy}{dx} = \sqrt{3} \).

Teacher's Note:
a) Use implicit differentiation carefully, applying the product rule to the xy term.
b) Substitute the given point only after isolating dy/dx.

23. (a) Simplify : \( \sin^{-1}\sqrt{\dfrac{1+\cos 2x}{2}} \), \( 0 \lt x \lt \dfrac{\pi}{2} \). [2 Marks]

Answer:
1. \( \dfrac{1+\cos 2x}{2} = \cos^2 x \), so the expression becomes \( \sin^{-1}(\cos x) \).
2. \( \sin^{-1}(\cos x) = \sin^{-1}\left(\sin\left(\dfrac{\pi}{2}-x\right)\right) = \dfrac{\pi}{2}-x \).

Teacher's Note:
a) Use the half angle identity \( \cos^2 x = \dfrac{1+\cos 2x}{2} \).
b) Convert cos x to sin form using the complementary angle identity.

OR

(b) Evaluate : \( \cos[\sin^{-1}(-1) - \tan^{-1}(-\sqrt{3})] \). [2 Marks]

Answer:
1. \( \sin^{-1}(-1) = -\dfrac{\pi}{2} \) and \( \tan^{-1}(-\sqrt{3}) = -\dfrac{\pi}{3} \).
2. So the expression is \( \cos\left(-\dfrac{\pi}{2}+\dfrac{\pi}{3}\right) = \cos\left(-\dfrac{\pi}{6}\right) = \dfrac{\sqrt{3}}{2} \).

Teacher's Note:
a) Recall principal value ranges of \( \sin^{-1} \) and \( \tan^{-1} \).
b) Simplify the angle first before applying the cosine function.

24. Using vectors, find the area of \( \triangle ABC \) with vertices A(1, 2, 3), B(2, -1, 4) and C(4, 5, -1). [2 Marks]

Answer:
1. \( \vec{AB} = \hat{i}-3\hat{j}+\hat{k} \) and \( \vec{AC} = 3\hat{i}+3\hat{j}-4\hat{k} \).
2. \( \vec{AB} \times \vec{AC} = 9\hat{i}+7\hat{j}+12\hat{k} \).
3. Area \( = \dfrac{1}{2}|\vec{AB}\times\vec{AC}| = \dfrac{1}{2}\sqrt{274} \) square units.

Teacher's Note:
a) Always find two sides from the same vertex before taking the cross product.
b) The area of a triangle is half the magnitude of the cross product of two sides.

25. Vectors \( \vec{a} = 3\hat{i}-2\hat{j}+2\hat{k} \) and \( \vec{b} = \hat{i}+2\hat{k} \) represent the two adjacent sides of a parallelogram. Find the vectors representing its diagonals and hence find their lengths. [2 Marks]

Answer:
1. One diagonal is \( \vec{a}+\vec{b} = 4\hat{i}-2\hat{j}+4\hat{k} \), with length \( \sqrt{16+4+16} = 6 \).
2. The other diagonal is \( \vec{a}-\vec{b} = 2\hat{i}-2\hat{j} \), with length \( \sqrt{4+4} = 2\sqrt{2} \).

Teacher's Note:
a) In a parallelogram with adjacent sides a and b, the diagonals are a+b and a-b.
b) Always compute the magnitude using the square root of the sum of squares of components.

SECTION C

This section comprises Short Answer (SA) type questions of 3 marks each.

26. Evaluate : \( \displaystyle\int_0^1 \log(1+x^2)\,dx \) [3 Marks]

Answer:
1. Using integration by parts, \( \displaystyle\int_0^1 \log(1+x^2)dx = [x\log(1+x^2)]_0^1 - \int_0^1 \dfrac{2x^2}{1+x^2}dx \).
2. This equals \( \log 2 - 2\displaystyle\int_0^1 \left(1-\dfrac{1}{1+x^2}\right)dx = \log 2 - 2[x-\tan^{-1}x]_0^1 \).
3. Evaluating the limits gives \( \log 2 - 2 + \dfrac{\pi}{2} \).

Teacher's Note:
a) Take \( \log(1+x^2) \) as the first function and 1 as the second in integration by parts.
b) Simplify the resulting rational expression by writing the numerator as \( (1+x^2)-1 \).

27. (a) Out of two bags, bag I contains 3 red and 4 white balls and bag II contains 8 red and 6 white balls. A die is thrown. If it shows a number less than 3 then a ball is drawn at random from bag I, otherwise a ball is drawn at random from bag II. Find the probability that the ball drawn from one of the bags is a red ball. [3 Marks]

Answer:
1. Let \( E_1 \): number \( \lt 3 \), so \( P(E_1)=\dfrac{2}{6}=\dfrac{1}{3} \); \( E_2 \): number \( \ge 3 \), so \( P(E_2)=\dfrac{4}{6}=\dfrac{2}{3} \).
2. \( P(\text{Red}|E_1)=\dfrac{3}{7} \) and \( P(\text{Red}|E_2)=\dfrac{8}{14}=\dfrac{4}{7} \).
3. Required probability \( = \dfrac{1}{3}\times\dfrac{3}{7} + \dfrac{2}{3}\times\dfrac{4}{7} = \dfrac{1}{7}+\dfrac{8}{21} = \dfrac{11}{21} \).

Teacher's Note:
a) Use the law of total probability with the die outcome as the conditioning event.
b) Carefully identify which bag corresponds to each die outcome range.

OR

(b) The probability of simultaneous occurrence of atleast one of the two events X and Y is a. If the probability that exactly one of the events X, Y occurs is b, prove that P(X') + P(Y') = 2 - 2a + b. [3 Marks]

Answer:
1. Given \( P(X\cup Y)=a \) and \( P(X\cup Y)-P(X\cap Y)=b \), so \( P(X\cap Y)=a-b \).
2. \( P(X')+P(Y') = [1-P(X)]+[1-P(Y)] = 2-[P(X)+P(Y)] \).
3. Using \( P(X)+P(Y)=P(X\cup Y)+P(X\cap Y)=a+(a-b) \), we get \( P(X')+P(Y')=2-2a+b \).

Teacher's Note:
a) Recall that exactly one event occurring means \( P(X\cup Y) - P(X\cap Y) \).
b) Use the addition theorem \( P(X)+P(Y) = P(X\cup Y)+P(X\cap Y) \).

28. (a) Find \( \displaystyle\int \sqrt{\dfrac{x+2}{x-2}}\,dx \) [3 Marks]

Answer:
1. Multiplying by \( \dfrac{\sqrt{x+2}}{\sqrt{x+2}} \) gives \( \displaystyle\int \dfrac{x+2}{\sqrt{x^2-4}}\,dx \).
2. Split as \( \dfrac{1}{2}\displaystyle\int \dfrac{2x}{\sqrt{x^2-4}}dx + 2\displaystyle\int\dfrac{1}{\sqrt{x^2-4}}dx \).
3. The first part gives \( \sqrt{x^2-4} \) and the second gives \( 2\log|x+\sqrt{x^2-4}| \).
4. So the integral \( = \sqrt{x^2-4} + 2\log|x+\sqrt{x^2-4}| + C \).

Teacher's Note:
a) Rationalising the square root is the key first step in this type of integral.
b) Split the numerator into a term proportional to the derivative of \( x^2-4 \) and a constant.

OR

(b) Find : \( \displaystyle\int \dfrac{x^2}{(x^2+9)(x^2+16)}\,dx \) [3 Marks]

Answer:
1. Put \( t=x^2 \) so \( \dfrac{t}{(t+9)(t+16)} = \dfrac{A}{t+9}+\dfrac{B}{t+16} \), giving \( A=-\dfrac{9}{7}, B=\dfrac{16}{7} \).
2. So the integral becomes \( -\dfrac{9}{7}\displaystyle\int\dfrac{1}{x^2+9}dx + \dfrac{16}{7}\displaystyle\int\dfrac{1}{x^2+16}dx \).
3. This equals \( -\dfrac{3}{7}\tan^{-1}\left(\dfrac{x}{3}\right) + \dfrac{4}{7}\tan^{-1}\left(\dfrac{x}{4}\right) + C \).

Teacher's Note:
a) Use partial fractions in terms of \( t=x^2 \) before integrating.
b) Remember the standard formula \( \int \dfrac{dx}{x^2+a^2}=\dfrac{1}{a}\tan^{-1}\dfrac{x}{a} \).

29. If \( I_1 = \displaystyle\int_{-\pi/4}^{\pi/4} \dfrac{dx}{1+\cos 2x} \) and \( I_2 = \displaystyle\int_{-1/2}^{1/2} |x|\,dx \), then show that \( I_1 - 4I_2 = 0 \). [3 Marks]

Answer:
1. Since \( \dfrac{1}{1+\cos 2x} \) is even, \( I_1 = 2\displaystyle\int_0^{\pi/4}\dfrac{dx}{2\cos^2 x} = \displaystyle\int_0^{\pi/4}\sec^2 x\,dx = [\tan x]_0^{\pi/4} = 1 \).
2. Since \( |x| \) is even, \( I_2 = 2\displaystyle\int_0^{1/2} x\,dx = [x^2]_0^{1/2} = \dfrac{1}{4} \).
3. Therefore \( I_1 - 4I_2 = 1 - 4\times\dfrac{1}{4} = 0 \).

Teacher's Note:
a) Use the property \( \int_{-a}^a f(x)dx = 2\int_0^a f(x)dx \) for even functions to simplify limits.
b) Simplify \( 1+\cos 2x = 2\cos^2 x \) before integrating.

30. (a) Find the general solution of the differential equation \( (y^2-x^2)dx = 2xy\,dy \) [3 Marks]

Answer:
1. Rewrite as \( \dfrac{dy}{dx} = \dfrac{y^2-x^2}{2xy} \), a homogeneous equation; put \( y=vx \), so \( \dfrac{dy}{dx}=v+x\dfrac{dv}{dx} \).
2. This gives \( x\dfrac{dv}{dx} = \dfrac{-(1+v^2)}{2v} \), so \( \dfrac{2v\,dv}{1+v^2} = -\dfrac{dx}{x} \).
3. Integrating gives \( \log(1+v^2) = -\log|x|+\log C \), i.e. \( (1+v^2)x=C \).
4. Substituting back \( v=\dfrac{y}{x} \) gives the general solution \( x^2+y^2 = Cx \).

Teacher's Note:
a) Recognise the equation as homogeneous since both terms have the same degree.
b) Always substitute back \( v = y/x \) at the end to express the solution in x and y.

OR

(b) Find the particular solution of the differential equation \( (1+e^{2x})dy + (1+y^2)e^x\,dx = 0 \), given that y(1) = 0. [3 Marks]

Answer:
1. Separating variables: \( \dfrac{dy}{1+y^2} = -\dfrac{e^x\,dx}{1+e^{2x}} \).
2. Integrating (using \( e^x=t \)) gives \( \tan^{-1}y + \tan^{-1}(e^x) = C \).
3. Using \( y(1)=0 \), \( C = \tan^{-1}(e) \), so the particular solution is \( \tan^{-1}y + \tan^{-1}(e^x) = \tan^{-1}(e) \).

Teacher's Note:
a) Separate the variables first, then substitute \( e^x=t \) for the right-hand integral.
b) Use the given initial condition to find the constant of integration.

31. Solve the following linear programming problem graphically : [3 Marks]
Minimize Z = 13x - 15y
Subject to constraints
x + y \( \le \) 7,
2x - 3y + 6 \( \ge \) 0,
x \( \ge \) 0, y \( \ge \) 0

[Figure: The feasible region is bounded by the lines x+y=7 and 2x-3y+6=0 together with the axes, with corner points at (0,0), (7,0), (3,4) and (0,2).]

Answer:
1. The corner points of the feasible region are (0,0), (7,0), (3,4) and (0,2).
2. Z at these points: Z(0,0)=0, Z(7,0)=91, Z(3,4)=-21, Z(0,2)=-30.
3. The minimum value of Z is -30, attained at x=0, y=2.

Teacher's Note:
a) Plot both constraint lines accurately and shade the common feasible region.
b) Always evaluate the objective function at every corner point to find the minimum.

SECTION D

This section comprises Long Answer (LA) type questions of 5 marks each.

32. Show that line AB passing through points A(0, 4, 1), B(2, 3, -1) and the line CD passing through points C(4, 5, 0), D(2, 6, 2) are parallel. Also, find distance between them. [5 Marks]

Answer:
1. Direction ratios of AB are \( \langle 2,-1,-2\rangle \), and of CD are \( \langle -2,1,2\rangle \), which are proportional, so the lines are parallel.
2. In vector form, \( \vec{r}=4\hat{j}+\hat{k}+\lambda(2\hat{i}-\hat{j}-2\hat{k}) \) and \( \vec{r}=4\hat{i}+5\hat{j}+\mu(2\hat{i}-\hat{j}-2\hat{k}) \).
3. \( \vec{a_2}-\vec{a_1} = 4\hat{i}+\hat{j}-\hat{k} \) and \( \vec{b}=2\hat{i}-\hat{j}-2\hat{k} \).
4. \( (\vec{a_2}-\vec{a_1})\times\vec{b} = -3\hat{i}+6\hat{j}-6\hat{k} \), with magnitude \( \sqrt{81}=9 \), and \( |\vec{b}|=3 \).
5. Distance between the lines \( = \dfrac{9}{3} = 3 \) units.

Teacher's Note:
a) First confirm parallelism by checking proportionality of direction ratios.
b) Use the formula \( d = \dfrac{|(\vec{a_2}-\vec{a_1})\times\vec{b}|}{|\vec{b}|} \) for distance between parallel lines.

33. (a) A relation R is defined on Z, the set of integers, as R = {(x, y) : |x - y| is divisible by a prime number 'p', x, y \( \in \) Z} check whether R is an equivalence relation or not. [5 Marks]

Answer:
1. Reflexive: For any \( x \in Z \), \( |x-x|=0 \), which is divisible by any prime p, so \( (x,x)\in R \).
2. Symmetric: If \( (x,y)\in R \), \( |x-y| \) is divisible by p; since \( |x-y|=|y-x| \), \( (y,x)\in R \).
3. Transitive: If \( (x,y),(y,z)\in R \), then \( x-y \) and \( y-z \) are divisible by p, so their sum \( x-z \) is also divisible by p, giving \( (x,z)\in R \).
4. Since R is reflexive, symmetric and transitive, R is an equivalence relation.

Teacher's Note:
a) Check all three properties separately and clearly with justification.
b) Use the fact that a sum of two multiples of p is again a multiple of p for transitivity.

OR

(b) A function f : \( R - \left\{\dfrac{3}{5}\right\} \longrightarrow R - \left\{\dfrac{3}{5}\right\} \) is defined as \( f(x) = \dfrac{3x+2}{5x-3} \). Show that f is one-one and onto. [5 Marks]

Answer:
1. Let \( f(x_1)=f(x_2) \): \( \dfrac{3x_1+2}{5x_1-3}=\dfrac{3x_2+2}{5x_2-3} \).
2. Cross-multiplying and simplifying gives \( x_1=x_2 \), so f is one-one.
3. For onto, let \( y \in R-\left\{\dfrac{3}{5}\right\} \) and solve \( f(x)=y \) to get \( x=\dfrac{3y+2}{5y-3} \), which lies in the domain.
4. Since for every y in the codomain there exists such an x, f is onto.

Teacher's Note:
a) Cross-multiply carefully and cancel common terms to prove one-one.
b) To prove onto, express x explicitly in terms of y and verify it lies in the domain.

34. (a) If A = \( \begin{bmatrix}0 & 2 & 1\\-2 & -1 & -2\\1 & -1 & 0\end{bmatrix} \), find A^-1 and use it to solve the following system of equations : -2y + z = 7, 2x - y - z = 8, x - 2y = 10 [5 Marks]

Answer:
1. \( |A| = -2(0-2)+1(2+1) = -4+3 = -1 \ne 0 \), so A is invertible.
2. \( \text{adj}A = \begin{bmatrix}-2 & -1 & -3\\-2 & -1 & -2\\3 & 2 & 4\end{bmatrix} \), so \( A^{-1} = \dfrac{1}{|A|}\text{adj}A = \begin{bmatrix}2 & 1 & 3\\2 & 1 & 2\\-3 & -2 & -4\end{bmatrix} \).
3. The system corresponds to \( A^TX=B \), where \( B=\begin{bmatrix}7\\8\\10\end{bmatrix} \), so \( X=(A^{-1})^TB \).
4. Computing gives \( x=0, y=-5, z=-3 \).

Teacher's Note:
a) Carefully identify that the coefficient matrix of the system is \( A^T \), not A itself.
b) Verify the inverse by checking \( |A| \ne 0 \) before proceeding.

OR

(b) If \( \begin{bmatrix}3 & -1 & \sin 3x\\-7 & 4 & \cos 2x\\-11 & 7 & 2\end{bmatrix} \) is a singular matrix, then find all values of x where \( x \in \left[0,\dfrac{\pi}{2}\right] \). [5 Marks]

Answer:
1. Since the matrix is singular, its determinant is 0: expanding gives \( 10 - 10\cos 2x - 5\sin 3x = 0 \).
2. Simplifying using \( \cos 2x = 1-2\sin^2x \) and \( \sin 3x = 3\sin x - 4\sin^3x \) gives \( 4\sin^3x+4\sin^2x-3\sin x=0 \).
3. Factorising: \( \sin x(2\sin x-1)(2\sin x+3)=0 \).
4. Since \( 2\sin x+3\ne 0 \), the solutions are \( \sin x=0 \) or \( \sin x=\dfrac{1}{2} \), giving \( x=0 \) or \( x=\dfrac{\pi}{6} \).

Teacher's Note:
a) Expand the determinant along the row or column with the trigonometric entries.
b) Convert all terms to a single trigonometric function (sin x) before factorising.

35. If x = a(sin t - t cos t) and y = b(cos t + t sin t), then find \( \dfrac{dy}{dx} \) and \( \dfrac{d^2y}{dx^2} \). [5 Marks]

Answer:
1. \( \dfrac{dx}{dt} = a(\cos t + t\sin t - \cos t) = at\sin t \).
2. \( \dfrac{dy}{dt} = b(-\sin t + t\cos t + \sin t) = bt\cos t \).
3. \( \dfrac{dy}{dx} = \dfrac{bt\cos t}{at\sin t} = \dfrac{b}{a}\cot t \).
4. Differentiating again, \( \dfrac{d^2y}{dx^2} = -\dfrac{b}{a}\text{cosec}^2t \times \dfrac{dt}{dx} = -\dfrac{b\,\text{cosec}^3t}{a^2t} \).

Teacher's Note:
a) Differentiate x and y separately with respect to t first.
b) For the second derivative, differentiate dy/dx with respect to t and multiply by dt/dx.

SECTION E

This section comprises of 3 case study based questions of 4 marks each.

36. Roundabouts are often made on busy roads to ease the traffic and avoid red lights. One such round-about is made such that equation representing its boundary is given by C1 ; x² + y² = 64. There is a circular pond with a fountain in the middle of the roundabout whose equation is given by C2 : x² + y² = 4. Based on the given information, answer the following questions :

[Figure: An aerial illustration of a circular roundabout with a fountain and pond at its centre, surrounded by trees.]

(i) Represent the given equations C1 and C2 with the help of a diagram. [1 Mark]

[Figure: Two concentric circles centred at the origin, the outer circle C1 of radius 8 and the inner circle C2 of radius 2.]

Answer: C1 is a circle of radius 8 centred at the origin, and C2 is a circle of radius 2 centred at the origin, with C2 lying entirely inside C1.

Teacher's Note:
a) Both circles are centred at the origin since there is no shift in the equations.
b) The radius is obtained by comparing with \( x^2+y^2=r^2 \).

(ii) Express y as a function of x, (y = f(x)), for both C1 an C2. [1 Mark]

Answer: For C1, \( y=\sqrt{64-x^2} \) or \( y=-\sqrt{64-x^2} \); for C2, \( y=\sqrt{4-x^2} \) or \( y=-\sqrt{4-x^2} \).

Teacher's Note:
a) Isolate y from the circle equation \( x^2+y^2=r^2 \) to get two branches.
b) The positive branch represents the upper semicircle and the negative branch the lower semicircle.

(iii) (a) Using integration find the area of region covered by the roundabout. [2 Marks]

Answer:
1. Required area \( = 4\displaystyle\int_0^8 \sqrt{64-x^2}\,dx \).
2. \( = 4\left[\dfrac{x}{2}\sqrt{64-x^2}+32\sin^{-1}\dfrac{x}{8}\right]_0^8 = 4\times16\pi = 64\pi \) square units.

Teacher's Note:
a) Use the standard integral formula \( \int\sqrt{a^2-x^2}dx = \dfrac{x}{2}\sqrt{a^2-x^2}+\dfrac{a^2}{2}\sin^{-1}\dfrac{x}{a} \).
b) Multiply by 4 to cover all four quadrants of the circle.

OR

(iii) (b) Using integration, find the area of region covered by circular pond. [2 Marks]

Answer:
1. Required area \( = 4\displaystyle\int_0^2 \sqrt{4-x^2}\,dx \).
2. \( = 4\left[\dfrac{x}{2}\sqrt{4-x^2}+2\sin^{-1}\dfrac{x}{2}\right]_0^2 = 4\times\pi = 4\pi \) square units.

Teacher's Note:
a) This is the standard area formula applied with radius 2.
b) The result should match \( \pi r^2 \) as a quick check, here \( \pi\times2^2=4\pi \).

37. An online delivery company in a city has 5000 subscribers and collects annual subscription fees of Rs. 300 per subscriber for unlimited free deliveries. The company wishes to increase the annual subscription fee. It is predicted that, for every increase of Rs. 1, ten subscribers will discontinue. Assume that the company increased the annual fee by Rs. x. Based on the given information, answer the following questions :

[Figure: An illustration of an online shopping delivery scene beside a graph showing number of subscribers decreasing as subscription fee increases.]

(i) How many subscribers will discontinue after an increase of Rs. x in annual fee ? [1 Mark]

Answer: 10x subscribers will discontinue.

Teacher's Note:
a) Each Re.1 increase causes 10 subscribers to leave, so for Rs. x increase, it is \( 10x \).
b) This is a direct proportional relationship given in the problem.

(ii) If R(x) denotes the total revenue collected after the increase of Rs. x in subscription fee, express R(x) as a function of x. [1 Mark]

Answer: \( R(x) = (5000-10x)(300+x) \).

Teacher's Note:
a) Revenue = (number of remaining subscribers) x (new fee per subscriber).
b) Remaining subscribers are \( 5000-10x \) and the new fee is \( 300+x \).

(iii) (a) Find the value of x for which R(x) is maximum. [2 Marks]

Answer:
1. \( R(x)=1500000+2000x-10x^2 \), so \( R'(x)=2000-20x \).
2. Setting \( R'(x)=0 \) gives \( x=100 \); since \( R''(x)=-20\lt0 \), R(x) is maximum at \( x=100 \).

Teacher's Note:
a) Expand R(x) first to make differentiation straightforward.
b) Use the second derivative test to confirm a maximum, not a minimum.

OR

(iii) (b) Find the sub-intervals of (0, 5000) in which R(x) is increasing and decreasing. [2 Marks]

Answer:
1. \( R'(x)=2000-20x \), which is zero at \( x=100 \).
2. For \( x\in(0,100) \), \( R'(x)\gt0 \), so R(x) is increasing on (0, 100); for \( x\in(100,5000) \), \( R'(x)\lt0 \), so R(x) is decreasing on (100, 5000).

Teacher's Note:
a) Test the sign of \( R'(x) \) on either side of the critical point x = 100.
b) State the intervals of increase and decrease clearly with respect to x = 100.

38. In an online jackpot, there is one first prize of Rs. 3,00,000, two second prizes of Rs. 2,00,000 each and three third prizes of Rs. 50,000 each. A total of 1,00,000 jackpot tickets each costing Rs. 100 were sold there by raising a fund of Rs. 1,00,00,000. Rohan bought one ticket. Based on given information, answer the following questions :

[Figure: An illustration of a "Lucky Ticket" showing prize amounts of Rs. 2,00,000 and Rs. 1,00,000 under a "WIN" banner.]

(i) What are the possible amounts, the person can win ? [1 Mark]

Answer: The person can win Rs. 3,00,000, Rs. 2,00,000, or Rs. 50,000 (or win nothing).

Teacher's Note:
a) List all distinct prize amounts as given in the problem.
b) Remember that not winning any prize is also a possible outcome.

(ii) (a) What is the probability that the person wins atleast Rs. 2,00,000 ? [2 Marks]

Answer:
1. Winning at least Rs. 2,00,000 means winning the first prize (1 ticket) or a second prize (2 tickets), total 3 favourable tickets out of 1,00,000.
2. Required probability \( = \dfrac{3}{1,00,000} \).

Teacher's Note:
a) Add the number of tickets for all prizes that are Rs. 2,00,000 or more.
b) Divide by the total number of tickets sold to get the probability.

OR

(ii) (b) What is the probability that the person does not win any amount ? [2 Marks]

Answer:
1. Total winning tickets = 1 + 2 + 3 = 6 out of 1,00,000.
2. Probability of not winning \( = 1 - \dfrac{6}{1,00,000} = \dfrac{99994}{1,00,000} = \dfrac{49997}{50000} \).

Teacher's Note:
a) First find the total number of prize-winning tickets by adding all categories.
b) Subtract the probability of winning from 1 to get the probability of not winning.

(iii) In another jackpot, Rohan also bought a ticket having a prize money of Rs. 5,00,000. The chances of winning the jackpot are 1 in 1,00,000. Find the probability that on exactly one of tickets he wins the jackpot. [1 Mark]

Answer: Required probability \( = \dfrac{6}{1,00,000}\times\dfrac{99999}{1,00,000} + \dfrac{99994}{1,00,000}\times\dfrac{1}{1,00,000} = \dfrac{699988}{10^{10}} \) (or equivalently \( \dfrac{174997}{2500000000} \)).

Teacher's Note:
a) "Exactly one" means either the first ticket wins and the second loses, or the first loses and the second wins.
b) Add the two mutually exclusive probabilities to get the final answer.

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